## Calculation of Carbon to Nitrogen Ratio (C:N)

This guide is for those of you out there that want to calculate the Carbon to Nitrogen Ratio, this is a simplified version. We have a more simplified version in our creating a compost heap guide.

Organic matter is not solely carbon or solely nitrogen. All live organic matter has some of both elements, in varying proportions. After death, the ratio of carbon to nitrogen increases over time.

The following examples explain how to compute the recommended 30:1 ratio of Carbon to Nitrogen in a compost pile. Most composters use trial and error to attain a good ratio, but it is helpful to some to understand the mathematical concept.

If you don’t want to understand the mathematical concept, don’t worry about it.

Please be aware that our formula is simplified for use by home composters. To be truly correct, you must factor in the moisture content, bioavailability of carbon and nitrogen, and lignin content in the materials. We will instead use the simpler method of considering estimated ratios of carbon to nitrogen.

For purposes of this example, we are going to round the grass ratio up to 20:1, and use 40:1 for the leaves ratio. We will also use rounding frequently in performing the math. These equations illustrate how to set up the problems.

The examples given are:

- Example of determining C:N ratio (2-input pile)
- Example of determining materials needed by weight (2-input pile)
- Example of determining C:N ratio (3-input pile)
- Example of determining materials needed by weight (3-input pile)

### Example of determining C:N ratio (2-input pile):

You have 5 pounds of grass clippings (20:1).

You have 5 pounds of leaves (40:1).

You have a total of 10 lbs. of material: 50% are grass, 50% are leaves.

Multiply the % of grass by the C:N ratio of grass, add the multiplication of the % of leaves by the C:N ratio of leaves.

(50% × 20/1) + (50% × 40/1)=

10 + 20 =

30 —> which in fraction notation is 30/1 or 30:1.

The C:N ratio is 30:1.

### Example of determining materials needed by weight (2-input pile):

You have 5 pounds of rotted manure (25:1) and you want to know how much corn stalks (60:1) in weight to add to get the optimum 30:1 ratio.

Set the unknown variable, weight of cornstalks needed, to equal “W”.

Set the total weight of the pile equal to “T”.

The % of the total pile by weight represented by cornstalks will be = W/T.

Therefore, the weight of the rotted manure will equal (T – W) = 5 lbs.

The % of the total pile by weight represented by manure will be 5/T.

Fill these variables and the known 30:1 target ratio into the equation from the first example. In other words, multiply the % of manure by the C:N ratio of manure, add the multiplication of the % of cornstalks by the C:N ratio of cornstalks, equal to a ratio of 30/1.

(5/T × 25/1) + (W/T × 60/1) = 30/1 —> Perform basic math to reduce the equation.

125/T + 60W/T = 30

125+60W=30T

since T = (5 + W), 125 + 60W = 150 + 30W

30W = 25

W = 25/30 = .83 lbs. of cornstalks required

To check, put back into original equation:

W = .83, T = 5.83, Percent of total weight in manure = 5/5.83 = 86%, 100-86 = 14% for cornstalks.

(86% × 25/1) + (14% × 60/1) =

21.5 + 8.4 = 29.9 (not 30 due to rounding)

### Example of determining C:N ratio (3-input pile):

You have 5 pounds of grass clippings (20:1).

You have 5 pounds of leaves (40:1).

You have 2 pounds of rotted manure (25:1) You have a total of 12 lbs. of material. 42% is grass, 42% is leaves, 16% is manure.

Multiply the % of grass by the C:N ratio of grass, add the multiplication of the % of leaves by the C:N ratio of leaves, add the multiplication of the % of manure by the C:N ratio of manure.

(42% × 20/1) + (42% × 40/1) + (16% × 25/1)=

8.4 + 16.8 + 4 =

29.2 —> which in fraction notation is 29.2/1 or 29.2:1.

### Example of determining materials needed by weight (3-input pile):

You have 5 pounds of rotted manure (25:1) and 4 pounds of grass clippings (20:1) and you want to know how much corn stalks (60:1) in weight to add to it to get the optimum 30:1 ratio.

Set the unknown variable, weight of cornstalks needed, to equal “W”.

Set the total weight of the pile equal to “T”.

The % of the total pile by weight represented by cornstalks will be = W/T.

Therefore, the weight of the rotted manure will equal (T – W – 4) = 5 lbs.

The % of the total pile by weight represented by manure will be 5/T.

Therefore, the weight of the grass will equal (T – W – 5) = 4 lbs.

The % of the total pile by weight represented by grass will be 4/T.

Fill these variables and the known 30:1 target ratio into the equation used in the other examples. In other words, multiply the % of manure by the C:N ratio of manure, add the multiplication of the % of grass by the C:N ratio of grass, add the multiplication of the % of cornstalks by the C:N ratio of cornstalks, equal to a ratio of 30/1.

(5/T × 25/1) + (4/T × 20/1) + (W/T × 60/1) = 30/1 —>Perform basic math to reduce the equation.

125/T + 80/T + 60W/T = 30

205+60W=30T

since T = (5 + 4 + W), 205 + 60W = 270 + 30W

30W = 65

W = 65/30 = 2.2 lbs. of cornstalks required

To check, put back into original equation:

W = 2.2; T = 11.2; Percent of total weight in manure = 5/11.2 = 45%; Percent of total weight in grass = 4/11.2 = 36%; 100 – 45 – 36 = 11.2% for cornstalks.

(45% × 25/1) + (36% × 20/1) + (19% × 60/1) = =

11.3 + 7.2 + 11.4 = 29.9 (not 30 due to rounding)

2.2 lbs. of cornstalks are required